STRUCTURAL DESIGN FORMULAS

Finite-element-analysis (FEA) techniques, now common in plastic part design, provide valuable information about the mechanical performance of complex or critical designs. For simple geometries and noncritical parts, standard design formulas can give good results if the material remains within its elastic limit.

Even in a complex part, an area or feature under load can often be represented by standard formulas.

Because they are primarily a function of part geometry and load and not material properties, stress calculation formulas derived for metals apply directly to plastics. Generally material dependent, deflection formulas require elastic (Young’s) modulus and sometimes Poisson’s ratio, ν. Poisson’s ratio varies slightly with temperature and loading conditions, but usually only to an insignificant degree. Single-point data suffices for most calculations. Table 3-1 lists typical values for a variety of materials.

Use of Moduli

For short-term loads at room tempera-tures and stress levels below a resin’s proportional limit, use the instantaneous elastic modulus. At other temperatures, use isothermal stress-strain curves to calculate elastic modulus — simply stress divided by strain in the linear region — at the desired temperature.

Simple bending calculations involving The white line shows the suggested

design limit at various temperatures for a Bayblend PC/ABS resin used in applications subjected to dynamic fatigue loading for 10⁷cycles.

Fatigue properties are sensitive to many factors including notch effects, environ-mental factors, stress concentrators, loading frequency, and temperature.

Surface texture, surface finish, and whether the part is plated also affect fatigue performance. In contrast to metals, plastics have a high degree of inherent damping and relatively low thermal conductivity. Therefore, vibration frequencies as low as 10 Hz can cause heat generation in plastic parts. This can lead to thermal failure if the energy cannot be properly dissipated by other means, such as convection.

Fiber orientation can also affect fatigue performance. Fatigue strength for a given fiber-filled resin can be many times greater when the fibers are aligned lengthwise in the direction of loading rather than perpendicularly.

When calculating fatigue-life values, use fatigue data that is appropriate for your application, and always include a suitable safety factor.

solid plastics undergoing short-term loading below the proportional limit can use either the flexural modulus or the published instantaneous tensile modulus.

For short-term loads in the nonlinear region above the proportional limit, such as assembly stresses, you will have to use a secant modulus, calculated from the curves and based upon the actual calculated stress. To calculate secant modulus, first solve the stress equation, which is independent of the elastic modulus for the material. Next read the strain corresponding to this calculated stress on the appropriate stress-strain curve. Then, divide the calculated stress by the strain to obtain the secant modulus for that stress level.

The secant modulus typically provides satisfactory predictions of deflections in applications that experience higher strain levels. See example problem 3-3 for a demonstration of this procedure.

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Stress limits are best determined from isochronous stress-strain curves showing either crazing or design limits for the given time and temperature. Of course, appropriate safety factors should always be used. Use a safety factor of at least 2.0 — higher values are necessary in critical applications. General stress limits (such as 25% of the published tensile yield stress) usually have large inherent safety factors, but become less conserv-ative at elevated temperatures or long-time use conditions. To apply a stress limit, simply solve the stress equation for the For long-term loads, use a creep or

apparent modulus derived from isochro-nous stress-strain curves. A time-dependent property, creep modulus is the calculated stress divided by the corresponding strain value read from the isochronous stress-strain curve for the desired time span. Because the strain value is always changing in a part that is exhibiting creep, the creep modu-lus is also time dependent. Calculations using the creep modulus, a decreased-representative modulus value, predict the deflection that occurs after a period of time. See the Long-Term Properties section in this chapter for more infor-mation and example problems dealing with creep behavior.

Stress and Strain Limits

Plastics differ in the level of stress or strain they can tolerate in structural applications. Engineering strain is defined as the change in length of a specimen divided by its original length and is represented by the symbol ε. The actual units of strain are length divided by length (inches per inch, millimeters per millimeter) but it is most often represented as a percentage. Stress has units of force per cross-sectional area (pounds per square inch = psi, Newtons per square millimeter = Megapascals, MPa). Because stress and strain are interrelated, plastic parts can be designed based on either stress or strain limits.

given load and geometry to determine if the limit is exceeded. Be sure to multiply the result by an appropriate stress-concentration factor (see figure 3-32) before making the comparison. If the limit is exceeded, reduce the load or increase the cross-sectional area to reduce stress below the limit. Note that because the stress equation itself is not modulus-dependent, it is almost always used in conjunction with the deflection equation to evaluate true design performance.

Table 3-3 lists the permissible short-term strain limits at room temperature for various families of Bayer engineering plastics. One-time, short-duration load applications that stay below these limits typically do not fracture or exhibit significant permanent deformation.

Designs that see multiple applications of an applied load should stay below 60% of these values. Permissible strain values are typically used to design parts with short-term or intermittent loads such as cantilever snap arms. If a strain-based formula is not available, it can be created by substituting σ / ε for E in the deflection equation, then substituting the complete stress equation for σ.

Permissible Short-Term Strain Limits at 23°C (73°F) Table 3-3

General guide data for the allowable short-term strain for snap joints (single joining operation); for frequent separation and rejoining, use about 60% of these values.

Unreinforced

Apec High Heat PC 4.0%

Bayblend PC/ABS 2.5%

Centrex ASA 1.9%

Durethan PA cond. 6.0%

dry 4.0%

Lustran ABS 1.8%

Makroblend Polycarb. Blends 3.5%

Makrolon PC 4.0%

Triax PA/ABS 3.4%

Glass-Fiber-Reinforced (% Glass)

Makrolon (10%) PC 2.2%

Triax (15%) PA/ABS 2.2%

Makrolon (20%) PC 2.0%

Durethan (30%) PA cond. 2.0%

dry 1.5%

Example 3-1: Tensile Stress and Strain

A 5-inch-long bar with a cross section of 0.5 inch by 0.125 inch is exposed to a 250-pound tensile load. Calculate the stress and elongation of the Makrolon polycarbonate bar.

The definition of stress is load divided by cross-sectional area, so the stress is:

Note that no modulus values are required to determine the stress, simply load and cross-sectional area. (In some cases however, Poisson’s ratio is required.)

To find the elongation of the bar, deter-mine the strain (change in length per unit length) created by the applied 4,000-psi stress. Using Young’s modulus to calculate strain gives:

However, reading from the stress-strain curve at room temperature (23°C) in figure 3-2 gives a value of 1.35% strain for a stress of 4,000 psi. Since this strain value is greater than that calculat-ed with Young’s modulus, the sample must be strained beyond the proportion-al limit. The proper secant modulus for this case is then:

ε = σ / E = 4,000 psi / 350,000 psi

= 0.011 in/in = 1.1% strain σt= P / A = 250 / [(0.5)(0.125)]

= 4,000 psi Uniaxial Tensile and Compressive Stress

Because most plastic part failures are tensile failures and this failure mode is easy to test, the majority of the available stress-strain data were produced using tensile test methods. The compressive strength of plastic usually exceeds the tensile strength, but because it is more difficult to test, the compressive strength is usually assumed to equal the tensile strength, which is a conservative assumption.

Depending on geometry, excessive compressive stress may cause the part to buckle. Long, slender shapes are the most susceptible to this failure mode.

Consult a strength-of-materials textbook or engineering handbook for analytical buckling formulas.

The definition of engineering strain is

∆L / L, so to find the change in length,

∆L, multiply the original length of the sample by the strain. For the Young’s modulus case, ∆L = (5 inch)(0.011) = 0.055 inch. But the correct answer using the actual stress-strain curve is ∆L = (5 inch)(0.0135) = 0.068 inch. In this case, the error introduced by using Young’s modulus was about 19%.

Keep in mind that these calculations are assuming short-term loading. If the 4,000-psi stress is not removed after a short time, the material will creep causing strain to increase. Based on the set of isochronous curves shown in figure 3-10, crazing will occur after 6 x 10⁴ hours (about 6.8 years) at a stress level of 4,000 psi. Extrapolating from this data, we can see that applying a safety factor of 2.0 and keeping the stress below 2,000 psi reduces the risk of crazing during the life of most parts in unharsh environments.

E_secant= 4,000 psi / 0.0135

= 296,296 psi

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In this formula, M represents the bending moment applied to the beam. Bending moment can be defined as applied force times the distance to the point of interest.

For the simple cantilever shown in figure 3-14, the moment at the attachment point is the load times the length of the beam, or P times L. The common units of moment are pound-inches or Newton-meters. The distance from the neutral plane to the point of interest is repre-sented by c, and the moment of inertia of the cross section (not to be confused with bending moment) is represented by

σb= Mc I Bending and Flexural Stress

Bending or flexing a plastic part induces both tensile and compressive stresses through the cross section, as shown in figure 3-14. Bending creates tensile stresses on the convex side of the part and compressive stresses on the concave side. The neutral plane defines the plane of zero stress in which the stress magnitude switches from tensile to compressive. The stress distribution through the thickness of the part is defined by the formula:

capital letter I. The moment of inertia indicates resistance to bending and has units of length to the fourth power (inches⁴, millimeters⁴). Defining sec-tion modulus, Z (not to be confused with the material modulus, E) as I divided by c allows the bending-stress formula to be rewritten:

σb= M / Z

Figure 3-14 Tensile and Compressive Stresses in Bending

h c P L

Tensile Stress Neutral Plane Compressive Stress

σt(+)

σc(–)

Cross-Sectional Area Moment of Inertia Section Modulus

Shape A c I z = I/c

Section Properties for Bending Table 3-4

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For design purposes, the maximum tensile bending stress is of primary interest. The maximum tensile bending stress is found when c is set equal to the distance from the neutral plane to the outer surface in tension.

Table 3-4 shows formulas for the cross-sectional area, A; distance from the neutral plane to the outer surface in tension, c; moment of inertia, I; and section modulus, Z, for various cross sections. The dashed line in the cross-sectional diagrams denotes the neutral plane, or in this case, neutral axis. The formulas assume the bending moment is applied about this axis. The cross sections that are not symmetrical about the neu-tral axis require some back-substitution of A and c to calculate I and Z.

Bending-stress formulas are highly dependent on boundary conditions.

Boundary conditions define how the ends of the part are restrained, as well as the position of the load and whether it is concentrated or distributed across the surface of the part. Table 3-5 gives stress and deflection formulas for the bending of beams with different boundary conditions. The symbol P denotes con-centrated loads (pounds, Newtons) and the symbol w denotes loads evenly dis-tributed across the beam (pounds/inch, Newtons/millimeter). Use the values from table 3-4 for I and Z. For accurate results, use the secant modulus or apparent modulus for E.

Loading and Bending Stress Deflection

Boundary Conditions σb y

Beam Bending Formulas Table 3-5

The stress result is needed in this case only to calculate the proper secant modulus. Because the resin is 30% glass reinforced, fiber orientation is considered.

The gate at one end of the beam will align most of the fibers along the length of the beam, therefore, the curves in figure 3-5 apply. Reading from the 60°C curve at a stress of 4,354 psi (30 MPa) gives a strain of 1.3%. The secant modulus for this case is 4,354 psi / 0.013 = 334,923 psi.

Now solve the deflection equation using the secant modulus.

Example 3-2: Beam Bending

A load of 250 pounds is placed on a 10-inch-long beam 4 inches from one end. The I-shaped beam is 1-inch wide and 1-inch tall with a uniform thickness of 0.2 inch. The environmental tempera-ture is 140°F (60°C). The beam was injection molded from Durethan BKV 130 PA 6 resin through a gate on one end. Find the maximum deflection of the beam and at what point the maxi-mum deflection occurs.

First, calculate the section properties of the I-beam. From table 3-4 with b = h = 1, s = t = 0.2 and d = 0.6:

Now find the appropriate stress formula for the given boundary conditions in table 3-5. The fourth condition is correct.

Solving for maximum tensile bending stress with a = 6, b = 4, L = 10, and P = 50 gives:

σb= Pab / LZ

= (250)(6)(4) / [(10)(0.1378)]

= 4,354 psi (30 MPa) c = h / 2 = 1 / 2 = 0.5 inch

I = [(1)(1)³- (0.6)³(1-0.2)] / 12

= 0.0689 inch⁴

Z = I / c = 0.0689 / 0.5 = 0.1378 inch³

For this special case, the maximum deflection does not occur at the point where the load is applied. It instead occurs at:

x_m= [(L²-b²) / 3]^1/2= [(10²-4²) / 3]^1/2

= 5.29 inches from the left end of the beam

y =Pb(L²-b²)^3/2 15.6EIL

(250)(4)(10²-4²)^3/2

=(15.6)(334,923)(0.0689)(10)

= 0.214 inch

1.5 in

A A

t = 0.20 in

p = 275 psi

Section A-A

δ ^{= ?}

Figure 3-15 Simply Supported Plate

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Using the 40°C isothermal stress-strain curve in figure 3-2, a 1.75% strain is found to correspond to a stress of about 4,900 psi. Dividing stress by strain gives a secant modulus of 280,000 psi.

Solving the deflection equation using this modulus value gives:

Using the room-temperature flexural modulus (330,000 psi) instead of the secant modulus at 40°C would have predicted a deflection of 0.0206 inches, an error of 15%.

Shear Stress

In tensile or compressive loading, the load is applied perpendicular to the cross section of interest. Shear stress is calculated by considering the stress on the cross section that lies in-plane or parallel to the load. The most common example of shear stress is the shearing of a bolt or pin as shown in figure 3-16.

The load in the plates creates a shear stress on the cross section B-B equal to the load, P, divided by the cross-sectional area of the pin, A. Shear stress is denoted by the Greek letter τ. The units of shear stress (psi) are the same as for tensile or bending stress.

δmax= 3(275)(0.75)⁴[5-4(0.38)-(0.38)²] 16(280,000)(0.20)³

= 0.0243 inches

where:

p = applied pressure load (275 psi) r = plate radius (0.75 inches) ν = Poisson’s ratio (0.38) t = plate thickness (0.2 inches) E = modulus of elasticity in psi

This pressure load will cause strain in the disk to exceed the proportional limit.

In addition, the elevated-temperature condition rules out the use of the room-temperature Young’s modulus. Therefore, first calculate the appropriate secant modulus to use in the deflection formula.

Solving the stress equation yields:

σmax= 3(275)(0.75)²(3+0.38) / 8(0.20)²

= 4,902 psi Example 3-3: Plate Deflection

Assume that the simply supported plate shown in figure 3-15 has a diameter of 1.5 inches and a thickness of 0.2 inches.

A uniform load of 275 psi is applied in an ambient temperature of 104°F (40°C). Using the stress-strain curves for Makrolon polycarbonate resin, determine the deflection of the plate.

The maximum deflection (δ) and stress (σ) for this case can be calculated from the formulas:

Mold dump well

Figure 3-16 Shear Stress on a Pin

The strain produced in torsion is a shear strain,γ. It can be related to tensile strain using the approximate relation:

This equation is useful for converting permissible tensile-strain limits to permissible shear-strain limits. Lastly, for a circular cross section, the angle of twist in radians can be calculated given the shear strain and geometry by:

Example 3-4: Torsion of a Round Shaft

A 0.2-inch-diameter, 0.5-inch-long, Makrolon polycarbonate shaft is part of a torsional latch. A torque of 5 inch-pounds is applied to activate the latch.

Find the shear stress in the shaft and the resulting angle of twist.

The polar moment of inertia for the round cross section is:

For this case, c = d / 2, or 0.1 inch. The maximum shear stress in the shaft is then:

τ = Tc / J = (5)(0.1) / (0.000157)

= 3,185 psi

J = πd⁴/ 32 = (3.14)(0.2)⁴/ 32

= 0.000157 inch

ϕ = 2γL / d, (d = shaft diameter) γ ≈ (1+ν)ε

Torsion

Shear stress is the primary type of stress in parts that experience torsional or twisting loads. The stress formula for torsion is analogous to the bending-stress formula, σb= Mc / I. The bending moment is replaced with a twisting moment, T, and the moment of inertia is replaced by a polar moment of inertia, J. The distance c now represents the distance from the centroid of the section to the outer surface. This yields the following formula:

For the torsion problem the angle of twist,ϕ, is analogous to deflection. It is defined as:

In this expression, L is the length of the shaft and G is the shear modulus of the material. Assuming linear elasticity, the shear modulus can be approximated from the tensile modulus and Poisson’s ratio using the relation:

G≈ E

To find the angle of twist, we need G, and therefore E. Combining the relations for G and γ and replacing the moduli with their stress/strain definitions gives the relation: σ ≈ 2τ. This allows us to calculate secant modulus from the tensile stress-strain curve with a stress value of 2 times τ, or 6,370 psi. Using the 23°C curve in figure 3-2 gives a secant modulus of about 6,370 psi / 0.023 = 277,000 psi.

The calculated angle of twist is then:

Note that the conversion factor between radians and degrees is 180/π.

ϕ = TL / JG

= (5)(0.5) / [(0.000157)(100,362)]

= 0.159 radians

= 9.1 degrees G≈ Es/ [2(1+ν)]

≈ 277,000 / [2(1+0.38)]

≈ 100,362 psi

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Part Shape

In many applications, the overall part shape is the predominant design factor affecting part stiffness and load-carrying capabilities. Taking steps early in the design stage to select a good basic shape can avoid expensive and/or troublesome measures later in the product develop-ment to achieve the desired strength and stiffness. Selecting inherently stiffer shapes seldom adds significantly to the final part costs.

Take advantage of the design flexibility in the molding process to maximize the stiffness of your design. Consider crowns or corrugations for large surfaces.

Flat surfaces lack inherent stiffness.

In document Engineering Polymers. Part and Mold Design THERMOPLASTICS. A Design Guide (Page 59-68)